10.4 Fermi–Dirac Statistics
543
+
1
2
U
(μ)
f
(()(( − μ)
2 dd + . . .
= U(μ)L 0 + U
(μ)L 1 + U
(μ)L 2 + . . . ,
(10.4.39)
with the L i given, in general, by
L i ≡ −
∞
0
(( − μ)
i f
(() d .
(10.4.40)
For the time being we need only L 0 , L 1 and L 2 : their values can be shown to be 1,
0, and π 2 (k B T ) 2 /3, respectively, so that U can be written as
U = U(μ) +
π 2
6
(k B T )
2 U
(μ) + . . . .
(10.4.41)
By applying the same procedure to expression (10.4.27), i.e., carrying out an
integration by parts as in Eq. (10.4.36), but with
N ce (() =
8π
3
2m e
h 2
3
2
V V
3
2
(10.4.42)
replacing U(() in Eq. (10.4.39), we arrive at a similar expression for N, namely
N ce = N ce (μ) +
π 2
6
(k B T )
2 N
ce (μ) + . . . .
(10.4.43)
From this last result, together with expression (10.4.42) and its derivatives with
respect to , we obtain for N ce the expression
N ce =
8π
3
2m e
h 2
3
2
V μ
3
2
1 +
π 2
8
(βμ)
−2
+ . . .
.
(10.4.44)
If we now rearrange this result by multiplying both sides of the equation by
(3/8π )(h 2 /2m e )
3
2 /V , we obtain
N
V
3
8π
h 2
2m e
3
2
= μ
3
2
1 +
π 2
8
(βμ)
−2
+ . . .
,
(10.4.45)
which becomes, upon employing expression (10.4.22a) for the Fermi energy F ,
3
2
F = μ
3
2
1 +
π 2
8
(βμ)
−2
+ . . .
.
(10.4.46)